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Chapter 4: Problem 39

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int 2 x\left(1-x^{-3}\right) d x$$

Short Answer

Expert verified
The antiderivative is \(x^2 + \frac{2}{x} + C\).

Step by step solution

01

Expand the Integrand

First, let's expand the expression inside the integral. The original expression is \[ \int 2x(1-x^{-3}) \, dx \]Expanding the expression gives us:\[ \int (2x - 2x^{-2}) \, dx \]
02

Integrate Each Term Separately

Now, we integrate each term separately. 1. Integrate the first term, \(2x\):\[ \int 2x \, dx = x^2 \]2. Integrate the second term, \(-2x^{-2}\):\[ \int -2x^{-2} \, dx = \frac{2}{x} \] Thus, the integral becomes:\[ x^2 + \frac{2}{x} + C \]Where \(C\) is the constant of integration.
03

Verify by Differentiation

To verify the solution, differentiate it:\( \frac{d}{dx} [x^2 + \frac{2}{x} + C] \)1. Derivative of \(x^2\) is \(2x\).2. Derivative of \(\frac{2}{x}\) is \(-2x^{-2}\).3. Derivative of \(C\), a constant, is 0.Putting it all together:\[ 2x - 2x^{-2} \]This matches the original expression inside the integral, confirming our solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative
The concept of an antiderivative, often referred to as the primitive or an antiderivative function, is an essential notion in calculus. When we talk about finding an antiderivative, we are essentially looking for a function whose derivative is the given function. In our problem, we began with the expression \(2x(1-x^{-3})\), and by expanding and integrating, our goal was to find a function that when differentiated, would return the same expression we started with.

Here are a few key points to remember about antiderivatives:
  • An antiderivative is not unique; there are infinitely many antiderivatives for a given function because of the constant of integration \(C\).
  • The process of finding an antiderivative reverses differentiation; however, it requires adding the constant \(C\) because differentiation of a constant yields zero.
  • The strategy often involves rewriting the function into a simpler form, as we did by expanding \(2x(1-x^{-3})\).
By finding the most general antiderivative in the problem, it allows students to see the interplay between differentiation and integration.
Indefinite Integral
The indefinite integral, also known as the antiderivative, is a fundamental concept in integral calculus. It represents a family of functions that differ only by a constant. In our example, when we solved for the indefinite integral \( \int 2x(1-x^{-3})\, dx \), we found \(x^2 + \frac{2}{x} + C\). This expression can be understood as the set of all possible antiderivatives of the integrand.

Some important aspects of indefinite integrals include:
  • The indefinite integral symbol \(\int\) represents the process of integration.
  • The most general antiderivative will always include a constant of integration \(C\), accounting for all vertical shifts of the function.
  • In real-world applications, indefinite integrals can model accumulations, such as total distance from a velocity function over time.
To solve indefinite integrals, it's often key to simplify the integrand through expansion or substitution, as demonstrated in the exercise.
Differentiation
Differentiation is the process of finding the derivative of a function. It plays a pivotal role in verifying the results we obtain from integration. In this exercise, after finding the indefinite integral, we checked our solution by differentiating it to ensure it matched the original integrand \(2x(1-x^{-3})\).

Key points to understand differentiation:
  • Derivatives represent rates of change or the slope of a function at any given point.
  • When differentiating a sum, differentiate each term separately, as we did with \(x^2 + \frac{2}{x} + C\).
  • Verifying an integral by differentiation is a practical tool to check the correctness of your antiderivative solution.
Through this process, students can appreciate the connection and balance between differentiation and integration, which are inverse operations in calculus.

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Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the s-axis is $$ \frac{d s}{d t}=v=9.8 t-3 $$ i) Find the body's displacement over the time interval from \(t=1\) to \(t=3\) given that \(s=5\) when \(t=0\) ii) Find the body's displacement from \(t=1\) to \(t=3\) given that \(s=-2\) when \(t=0\) iii) Now find the body's displacement from \(t=1\) to \(t=3\) given that \(s=s_{0}\) when \(t=0\) b. Suppose that the position \(s\) of a body moving along a coordinate line is a differentiable function of time \(t .\) Is it true that once you know an antiderivative of the velocity function ds/dt you can find the body's displacement from \(t=a\) to \(t=b\) even if you do not know the body's exact position at either of those times? Give reasons for your answer.
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